SOLUTION: an investment club invested part of $10,000 at 9% annual interest and the rest at 8%. If the annual income from these investments was $860, how much was invested at 8%

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Question 339965: an investment club invested part of $10,000 at 9% annual interest and the rest at 8%. If the annual income from these investments was $860, how much was invested at 8%
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = dollars invested at 9%
Let b = dollars invested at 8%
given:
(1) a+%2B+b+=+10000
.09a+%2B+.08b+=+860
------------------
(2) 9a+%2B+8b+=+86000
Multiply both sides of (1) by 8 and
subtract (1) from (2)
(2) 9a+%2B+8b+=+86000
(1) -8a+-+8b+=+80000
a+=+6000
And, from (1)
(1) 6000+%2B+b+=+10000
b+=+4000
$6000 was invested at 9% and
$4000 was invested at 8%
check:
.09a+%2B+.08b+=+860
.09%2A6000+%2B+.08%2A4000+=+860
540+%2B+320+=+860
860+=+860
OK